A) \[\frac{1}{4}\]
B) 1
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
[d] \[y={{a}^{1-n}}{{x}^{n}}\] or \[\frac{dy}{dx}={{a}^{1-n}}n{{x}^{n-1}}\] \[sub-normal=\left| y\frac{dy}{dx} \right|=\left| y{{a}^{1-n}}n{{x}^{n-1}} \right|\] \[=\left| {{a}^{1-n}}{{x}^{n}}{{a}^{1-n}}n{{x}^{n-1}} \right|=\left| {{a}^{2-2n}}{{x}^{2n-1}} \right|\] which is constant if \[2n-1=0\] or \[n=1/2\].
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